Solve 1/frac{1{x}-1/x+10}=720

$$\frac{ 1 }{ \frac{ 1 }{ x } - \frac{ 1 }{ x+10 } } = 720$$

Solve for x

$$x=-90$$

$$x=80$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and x+10 is x\left(x+10\right). Multiply \frac{1}{x} times \frac{x+10}{x+10}. Multiply \frac{1}{x+10} times \frac{x}{x}.

$$\frac{1}{\frac{x+10}{x\left(x+10\right)}-\frac{x}{x\left(x+10\right)}}=720$$

Since \frac{x+10}{x\left(x+10\right)} and \frac{x}{x\left(x+10\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{x+10-x}{x\left(x+10\right)}}=720$$

Combine like terms in x+10-x.

$$\frac{1}{\frac{10}{x\left(x+10\right)}}=720$$

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Divide 1 by \frac{10}{x\left(x+10\right)} by multiplying 1 by the reciprocal of \frac{10}{x\left(x+10\right)}.

$$\frac{x\left(x+10\right)}{10}=720$$

Use the distributive property to multiply x by x+10.

$$\frac{x^{2}+10x}{10}=720$$

Divide each term of x^{2}+10x by 10 to get \frac{1}{10}x^{2}+x.

$$\frac{1}{10}x^{2}+x=720$$

Subtract 720 from both sides.

$$\frac{1}{10}x^{2}+x-720=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{10} for a, 1 for b, and -720 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{10}\left(-720\right)}}{2\times \frac{1}{10}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{10}\left(-720\right)}}{2\times \frac{1}{10}}$$

Multiply -4 times \frac{1}{10}.

$$x=\frac{-1±\sqrt{1-\frac{2}{5}\left(-720\right)}}{2\times \frac{1}{10}}$$

Multiply -\frac{2}{5} times -720.

$$x=\frac{-1±\sqrt{1+288}}{2\times \frac{1}{10}}$$

Add 1 to 288.

$$x=\frac{-1±\sqrt{289}}{2\times \frac{1}{10}}$$

Take the square root of 289.

$$x=\frac{-1±17}{2\times \frac{1}{10}}$$

Multiply 2 times \frac{1}{10}.

$$x=\frac{-1±17}{\frac{1}{5}}$$

Now solve the equation x=\frac{-1±17}{\frac{1}{5}} when ± is plus. Add -1 to 17.

$$x=\frac{16}{\frac{1}{5}}$$

Divide 16 by \frac{1}{5} by multiplying 16 by the reciprocal of \frac{1}{5}.

$$x=80$$

Now solve the equation x=\frac{-1±17}{\frac{1}{5}} when ± is minus. Subtract 17 from -1.

$$x=-\frac{18}{\frac{1}{5}}$$

Divide -18 by \frac{1}{5} by multiplying -18 by the reciprocal of \frac{1}{5}.

$$x=-90$$

The equation is now solved.

$$x=80 x=-90$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and x+10 is x\left(x+10\right). Multiply \frac{1}{x} times \frac{x+10}{x+10}. Multiply \frac{1}{x+10} times \frac{x}{x}.

$$\frac{1}{\frac{x+10}{x\left(x+10\right)}-\frac{x}{x\left(x+10\right)}}=720$$

Since \frac{x+10}{x\left(x+10\right)} and \frac{x}{x\left(x+10\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{x+10-x}{x\left(x+10\right)}}=720$$

Combine like terms in x+10-x.

$$\frac{1}{\frac{10}{x\left(x+10\right)}}=720$$

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Divide 1 by \frac{10}{x\left(x+10\right)} by multiplying 1 by the reciprocal of \frac{10}{x\left(x+10\right)}.

$$\frac{x\left(x+10\right)}{10}=720$$

Use the distributive property to multiply x by x+10.

$$\frac{x^{2}+10x}{10}=720$$

Divide each term of x^{2}+10x by 10 to get \frac{1}{10}x^{2}+x.

$$\frac{1}{10}x^{2}+x=720$$

Multiply both sides by 10.

$$\frac{\frac{1}{10}x^{2}+x}{\frac{1}{10}}=\frac{720}{\frac{1}{10}}$$

Dividing by \frac{1}{10} undoes the multiplication by \frac{1}{10}.

$$x^{2}+\frac{1}{\frac{1}{10}}x=\frac{720}{\frac{1}{10}}$$

Divide 1 by \frac{1}{10} by multiplying 1 by the reciprocal of \frac{1}{10}.

$$x^{2}+10x=\frac{720}{\frac{1}{10}}$$

Divide 720 by \frac{1}{10} by multiplying 720 by the reciprocal of \frac{1}{10}.

$$x^{2}+10x=7200$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=7200+5^{2}$$

Square 5.

$$x^{2}+10x+25=7200+25$$

Add 7200 to 25.

$$x^{2}+10x+25=7225$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=7225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7225}$$

Simplify.

$$x+5=85 x+5=-85$$

Subtract 5 from both sides of the equation.

$$x=80 x=-90$$